O Level Physics
Unit 2: Kinematics
1.
Scalar and Vector quantities
By definition
Type of Quantity
Scalar
Vector
Has magnitude only
Has magnitude and
direction.
Examples
Distance
Displacement
Speed
Velocity
Time
Acceleration
Mass
Weight
Displacement is distance in a specified direction.
2.
Note: When asked to find
a) a scalar quantity, only
the magnitude is
required (e.g. 70m).
b) a vector quantity, the
magnitude and
direction are required.
(e.g. 75m, 20° from the
North.
Kinematic Quantities
Time
Distance/Displacement
Speed/Velocity
-Final velocity
-Initial velocity
Acceleration
Symbol
t
s
v
v
u
a
SI Unit
s
m
m/s
m/s
m/s
m/s2
Speed and Velocity
Definition
Type of quantity
SI Unit
Formula
Note
4.
Speed
Distance travelled
per unit time.
Scalar
Velocity
Change in
displacement per
unit time.
Vector
m/s
The magnitude of the speed and
velocity of an object will differ if there is
a change in direction.
Two moving objects are said to have the same velocity when they have the same speed and
move in the same direction. In a linear motion, velocity can be positive or negative to specify
the direction of motion denoted as positive and negative. A positive velocity describes motion
in the denoted positive direction; and negative velocity in the denoted negative direction (or
back towards starting point)
Note: In negative velocity (e.g. -6m/s), the negative sign = motion opposite to the positive
direction denoted. However, its speed is 6m/s as speed is a scalar and does not involve
direction.
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[Physics] Unit 2: Kinematics
3.
1
Acceleration
5.
Definition
SI unit
Type of Quantity
Formula
Acceleration
Change in velocity per unit time.
m/s2
Vector
Acceleration =
a=
Note: The formula may be
changed to the following:
a) a =
b) v = u + a(∆t)
c) ∆t =
6.
Acceleration occurs when there is a change in an object’s velocity (change in speed and/or
direction).
7.
Positive acceleration:
Negative acceleration:
Object accelerates in direction of velocity.
Hence, velocity is increasing.
Object accelerates in direction opposite to velocity.
Hence, velocity is decreasing.
For a linear motion, the direction of the acceleration can be specified by denoting as positive
and negative.
Note: When objects travel in a non-linear motion, it undergoes positive acceleration.
Graphical Analysis of Motion
Deductions of graphs
Deductions
Displacement-time (disp-t) graph
 The gradient of the d-t graph
is equivalent to the speed OR
velocity. [See note]
 A st line with positive grad ⇒
uniform velocity; st line with
negative grad ⇒ uniform
velocity in opposite
direction.
 Curve ⇒ non-uniform
velocity (positive
acceleration)
 Grad of the tangent at a
point ⇒ instantaneous speed
OR velocity. [See note]
Note: The grads of dist-t ⇒ speed;
of dspmt-t ⇒ velocity. Distance &
speed are scalars, displacement &
velocity are vectors. Note that
dspmt-t graph involves direction
unlike dist-t.
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Velocity-time (vel-t) Graph
 Grad of vel-t graph ⇒
acceleration. Positive grad
⇒ positive acc; negative
grad ⇒ negative
acc/deceleration (moving in
opp direction)
 St line ⇒ uniform acc; curve
⇒ non-uniform acc
(positive)
 Grad of the tangent at a
point ⇒ instantaneous acc.
 Area under the graph ⇒
distance travelled OR
displacement [See note].
[Physics] Unit 2: Kinematics
8.
2
9.
Summary- Graphical Analysis of Motion
At rest
Displacement-time (disp-t) graph
Velocity-time (vel-t) Graph
dspmt/m
v/ms
-1
t/s
Uniform velocity
t/s
-1
dspmt/m
v/ms
t/s
t/s
*Uniform velocity
in opposite
direction
*Note: Not applicable for
distance-time graph.
*Note: Not applicable for speedtime graph.
-1
dspmt/m
v/ms
t/s
t/s
Uniform
acceleration
(just acceleration
for disp-t)
Uniform
deceleration
(just deceleration
for disp-t)
-1
dspmt/m
v/ms
t/s
t/s
-1
v/ms
dspmt/m
t/s
t/s
Increasing
acceleration
-1
v/ms
t/s
Decreasing
acceleration
-1
v/ms
t/s
-1
v/ms
t/s
Decreasing
deceleration
-1
v/ms
t/s
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[Physics] Unit 2: Kinematics
Increasing
deceleration
3
10.
Acceleration of Free Fall
During free fall when there is no air resistance, all objects falling under gravity experiences
constant acceleration (acceleration due to gravity). The acceleration due to gravity, g, for a
body close to Earth’s surface is a constant 10m/s2. Every one second, the object’s speed will
increase by 10m/s.
Acceleration due to gravity where air resistance is negligible is independent of mass and
surface area. All objects undergo the same constant acceleration.
11.
g=
is equivalent to
10 =
;
W=mg
is equivalent to
W=m(10)
, since the constant g is known.
Free fall- Downward motion
0s
Vel/ms-1
0
g/ms-2
10
10m
1s
10
10
Object accelerates due to gravity
at 10m/s2. In 1s, it travels 10m.
30m
2s
20
10
Object accelerates due to gravity
at 10m/s2. In 1s, it travels 20m
Dspmt
0m
50m
3s
0
0
At the moment when object was
dropped, it was at rest.
Object at rest. Velocity and
acceleration are both zero.
v/ms-1
Displacement/m
a/ms-2
10
t/s
t/s
t/s
[Physics] Unit 2: Kinematics
Area under velocity/speed time graph=
displacement/distance
dropped respectively.
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12.
Free fall- Upward and downward motion.
Pstn
A
Time
0s
Vel/ms-1
-20
g/ms-2
-10
Dspmt/m
0
B
1s
-10
-10
-20
C
2s
0
Non-zero
-30
D
3s
10
10
20
E
4s
20
10
0
F
5s
30
10
30
C
B
D
A
E
F
Velocity/ms-1
Displacement/m
t/s
a/ms-2
t/s
t/s
Speed/ms-1
Distance/m
t/s
t/s
[Physics] Unit 2: Kinematics
Area under velocity/speedtime graph =
displacement/distance
respectively.
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13.
Air resistance
Air resistance is a frictional force which opposes motion. It causes the acceleration of an object
in free fall to be lower than the acceleration of free fall.
(g of object falling with air resistance < 10m/s2)
An object falling under gravity with air resistance experiences decreasing acceleration (air
resistance increases with increasing velocity) until terminal velocity is reached.
14.
Air resistance is dependent on
a) Velocity: Air resistance increases with velocity till terminal velocity.
b) Surface area: Air resistance is directly proportional to the surface area of an object.
15.
Terminal velocity is the maximum constant velocity that can be reached where a = 0 and which
occurs when (weight = air resistance).
Terminal velocity is dependent on:
a) Mass of object: A heavier object will experience a higher terminal velocity than a
lighter object
b) Surface area of object: An object with a larger SA will experience lower terminal
velocity than one with a smaller SA.
16.
-1
R and W
v/ms-1 a/ms-2 dspmt/m
R=W
0
0
0
R1 < W
8
8
8
R2 < W
14
6
22
R1 < R2
3
R3 < W
18
4
30
R1 < R2 < R3
4
R=W
20
0
32
Note: W is a constant, R increases with velocity.
t/s
0
1
2
v/ms
t/s
[Physics] Unit 2: Kinematics
Decreasing acceleration due to
increasing velocity hence
increasing air resistance.
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6
Questions involving calculations- Unit 2: Dynamics
Speed and Velocity
1.
A car travels 7km north and then 3km west in 10mins. Calculate its
a) average speed; b) average velocity.
a) Ave speed =
= 60km/h
b) By Pythagoras’ Theorem,
AC =
Displacement ≈
7.6158km
C
3km
B
Ave velocity =
= 45.7km/h (3sf)
Bearing of C from A =
= 336.8° (1dp)
Ave velocity is 45.7km/h, 336.8° from the North.
7km
N
A
Acceleration
2.
The figure shows the displacement-time graph of an object.
a) Fill in the table below and sketch the corresponding velocity-time graph.
Time
Displacement
Velocity
0 to 4s
4 to 8s
8 to 16s
[Physics] Unit 2: Kinematics
b) Calculate
i)
the average speed of the object
ii)
the average velocity of the object
c) If the object decelerates uniformly to rest in 8s at the end of the 16s, calculate the
distance moved in this period.
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3.
A motorist, who saw the traffic light he was approaching turn red, was travelling at 15m/s. His
reaction time is 0.4s.
a) Given that the maximum deceleration of the car is 3.75m/s2, calculate the distance
travelled by the car before it comes to a complete stop.
Given that the motorist is 40m away from the junction which he stopped exactly at, calculate
b) the car’s deceleration.
c) the time taken for the car to stop, using your answer in (b).
Acceleration of Free Fall
4.
A brick falls from the top of the building and takes 4.0s to reach the ground. Calculate
a) the speed of the brick when it reaches the ground;
b) the height from which it falls.
a) g =
b) Height = distance travelled
= average speed x time
= (0 + 40)(4.0)
=80m
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[Physics] Unit 2: Kinematics
10 =
v = 10 x 4
= 40m/s
8
5.
A book was dropped from a window 9m above the ground. Calculate
a) the time taken for the book to hit the ground
b) the speed of the book when it hits the ground.
a) g =
10 =
Final speed = 10t
Distance = average speed x time
9 = (10t) x t
5t2 = 9
t = 1.34s
Alternative mtd for (a)
g=
10 =
Final speed = 10t
-1
v/ms
10t
t/s
t
Area under v-t graph = distance = vt = 5t2
5t2 = 9
t = 1.34s
b) Final speed = 10t
= 13.4m/s
6.
A man takes off from a spring board as shown in the
diagram. He jumps up into the air to reach the highest
point of his jump before falling downwards.
(Take air resistance as negligible).
1.02s
[Physics] Unit 2: Kinematics
a) It is given that that the man took 0.7s to reach the
highest point of his jump. Calculate his speed
when he leaves the spring board.
b) He takes another 1.02 before entering the water.
Calculate his speed when he enters the water.
c) Find the height of the springboard above the
water.
0.7s
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9